A) 36
B) 49
C) 0.49
D) 0.36
Correct Answer: B
Solution :
\[\begin{align} & \begin{matrix} {} & {{H}_{2}}+ & {{I}_{2}}\rightleftharpoons & 2HI \\ \text{initial}\,\text{conc}\text{.} & a & a & 0 \\ \text{conc}\text{.}\,\text{at}\,\text{equil}\text{.} & (a-x) & (a-x) & x=0.7 \\ \end{matrix} \\ & \begin{matrix} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=0.1 & \,\,\,\,=0.1 \\ \end{matrix} \\ \end{align}\] \[\therefore \,\,\,\,{{K}_{c}}\,=\frac{{{[HI]}^{2}}}{[{{H}_{2}}]\,[{{I}_{2}}]}\,=\frac{{{(0.7)}^{2}}}{(0.1)\,(0.1)}\] \[=\frac{0.49}{0.01}\,=49\]You need to login to perform this action.
You will be redirected in
3 sec